Is ${232001}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {232001}= &&{2}\cdot100000+ \\&&{3}\cdot10000+ \\&&{2}\cdot1000+ \\&&{0}\cdot100+ \\&&{0}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {232001}= &&{2}(99999+1)+ \\&&{3}(9999+1)+ \\&&{2}(999+1)+ \\&&{0}(99+1)+ \\&&{0}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {232001}= &&\gray{2\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{0\cdot9}+ \\&& {2}+{3}+{2}+{0}+{0}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${232001}$ is divisible by $9$ if ${ 2}+{3}+{2}+{0}+{0}+{1}$ is divisible by $9$ Add the digits of ${232001}$ $ {2}+{3}+{2}+{0}+{0}+{1} = {8} $ If ${8}$ is divisible by $9$ , then ${232001}$ must also be divisible by $9$ ${8}$ is not divisible by $9$, therefore ${232001}$ must not be divisible by $9$.